Answer to Question #97137 in Calculus for Anand

Question

Find the angle of intersection between the curves x²+2xy–y²+2àx = 0 and 3y³–2a²x–4a²y+a³ = 0 at the point (a,–a).

Answer 1

The angle between the curves on the plane at their common point $x_0=(a,-a)$ is the smallest of the two possible angles between the tangents to these curves at a given point.

$\tan\phi=\cfrac{f_2'(x_0)-f_1'(x_0)}{1 + f_1'(x_0)f_2'(x_0)}$ , where $\phi$ is the angle of intersection between the curves, $f_1(x)$ is the first curve, $f_2(x)$ is the second curve.

1) $(x^2+2xy-y^2+2ax)'=2x+2y+2xy'-2yy'+2a=0$. Here we have used formula $(xy)'=x'y+xy'=y+xy'$ .

$x+y+xy'-yy'+a=0$ ,

$y'(x-y)=-(x+y+a)$ ,

$y'=f_1'(x)=\cfrac{-(x+y+a)}{x-y}$ ,

$y'(x_0)=f_1'(x_0)=\cfrac{-(a-a+a)}{a-(-a)}=\cfrac{-a}{2a}=-\cfrac{1}{2}$ .

2) $(3y^3-2a^2x-4a^2y+a^3)'=9y^2y'-2a^2-4a^2y'=0$ ,

$y'(9y^2-4a^2)=2a^2$ ,

$y'=f_2'(x)=\cfrac{2a^2}{9y^2-4a^2}$ ,

$y'(x_0)=f_2'(x_0)=\cfrac{2a^2}{9a^2-4a^2}=\cfrac{2}{5}$ .

So, $\tan\phi=\cfrac{(\cfrac{2}{5})^2-(-\cfrac{1}{2})^2}{1+\cfrac{2}{5}(-\cfrac{1}{2})}=\cfrac{\cfrac{4}{25}-\cfrac{1}{4}}{1-\cfrac{1}{5}}=-\cfrac{9}{80}$ .

$\phi = arctan(-\cfrac{9}{80})=-6.41878673\degree$ . For this operation you can use an arctan calculator (https://www.rapidtables.com/calc/math/Arctan_Calculator.html).

Answer: the angle of intersection between the curves at the point $(a,-a)$ is $-6.41878673\degree$ .